⏩Heya,⏪
50 points ✌✌✌
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Please solve this Q25. ASAP:
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Answer for convenience:
(i.) (0,3-4√3)
(ii.) (0,3+4√3)
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▶Note:- PLEASE GIVE ONLY RELEVANT ANSWERS:)
Answers
Given, P(0, 2) is equidistant from A(3, k) and B(k, 5).
∴ AP = PB
⇒ AP2 = PB2
⇒ (3 – 0)2 + (k – 2)2 = (k – 0)2 + (5 – 2)2 [ Using Distance formula]
⇒ 9 + k2 – 4k + 4 = k2 + 9
⇒ – 4k + 4 = 0
⇒ 4k = 4
⇒ k = 1
Thus, the value of k is 1.
Hope it helps you.
Given vertices of equilateral triangle are A(-4,3) and B(4,3).
Let the coordinates of the third vertex be C(x,y).
Distance between A(-4,3) and B(4,3):
⇒ √[(4 + 4)^2 + (3 - 3)^2]
⇒ [8^2 + 0^2]
⇒ 64.
Distance between A(-4,3) and C(x,y):
⇒ √[(x + 4)^2 + (y - 3)^2]
⇒ [(x + 4)^2 + (y - 3)^2]
⇒ [x^2 + 16 + 8x + y^2 + 9 - 6y]
⇒ [x^2 + y^2 + 8x - 6y + 25] ------- (1)
Distance between B(4,3) and C(x,y):
⇒ √[(x - 4)^2 + (y - 3)^2]
⇒ [(x - 4)^2 + (y - 3)^2]
⇒ [x^2 + 16 - 8x + y^2 + 9 - 6y]
⇒ [x^2 + y^2 - 8x - 6y + 25] --------- (2)
Let's Equate AC = BC = > AC^2 = BC^2
⇒ x^2 + y^2 + 8x - 6y + 25 = x^2 + y^2 - 8x - 6y + 25
⇒ x^2 + y^2 + 8x - 6y + 25 - x^2 - y^2 + 8x + 6y - 25 = 0
⇒ 16x = 0
⇒ x = 0.
Substitute the value of x = 0 in (1) and equate it to 64, we get
⇒ [x^2 + y^2 + 8x - 6y + 25] = 64
⇒ [0 + y^2 - 6y + 25] = 64
⇒ y^2 - 6y + 25 = 64
⇒ y^2 - 6y - 39 = 0
Now,
Discriminant D = b^2 - 4ac.
= (-6)^2 - 4(1)(-39)
= 36 + 156
= 192.
The solutions for the quadratic equation is:
(a)
y = -b + √D/2a
= [-(-6) + √192]/2
= (6 + √192)/2
= (3 + 4√3)/2.
(b)
y = (-b - √D)/2a
= (-(-6) - √192)/2
= (6 - √192)/2
= (3 - 4√3)/2.
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(i)
The origin lies in the interior of the triangle.
The third vertex(x,y) = (0,3 - 4√3).
(ii)
The origin lies in the exterior of the triangle.
The third vertex(x,y) = (0,3 + 4√3).
Hope this helps!